Sums of Uniform Variables and Piecewise Densities

Why Uniform Sums Have Piecewise Densities

A uniform density is flat, but the density of a sum of uniform variables need not be. For two variables, values near the middle of the possible range can be formed in more ways than values near either endpoint. For three variables, the same changing overlap produces a piecewise quadratic density. These examples make the geometry of sum distributions especially visible.

Let \(U_1,U_2,\ldots\) be independent random variables, each with the Uniform(0,1) distribution. Their joint density is constant on the unit square or cube and zero elsewhere. The density of a sum is determined by how much of that square or cube contributes at each possible sum. Here we derive the two-variable and three-variable densities and use their pieces to calculate probabilities.

For independent variables with densities, the density of their sum is the convolution of those densities. In the cases below, if \(X\) has density \(f_X\) and \(U\) is an independent Uniform(0,1) variable, this formula becomes

where the integrand is understood to be zero outside the support of \(X\). The second expression follows by setting \(v=s-u\). Thus the integral only counts the part of the interval \([s-1,s]\) that overlaps the support of \(X\). Equivalently, in the first expression only \(u\) values for which \(s-u\) is in that support contribute.

Two Uniform Variables: The Triangular Density

Set \(S_2=U_1+U_2\). For a fixed sum \(s\), the eligible values of \(u=U_1\) must satisfy both \(0\leq u\leq1\) and \(0\leq s-u\leq1\). These inequalities give \( \max(0,s-1)\leq u\leq\min(1,s) \). Because both input densities equal 1 on \([0,1]\), the convolution integral is the length of this interval. Its endpoints change at \(s=1\), which is why the density has two pieces.

To derive the formula directly, intersect the eligible interval for \(u\) with \([0,1]\). If \(0<s<1\), the eligible interval is \([0,s]\), of length \(s\). If \(1<s<2\), it is \([s-1,1]\), of length \(2-s\). For \(s<0\) or \(s>2\), there are no eligible values. This proves the stated density almost everywhere, which is sufficient because densities that differ only on a set of probability zero describe the same distribution. \(\square\)

The density rises linearly from 0 to 1, then falls linearly back to 0. This is the triangular density. Its total integral is 1:

The two pieces also reflect the symmetry of the sum: replacing each \(U_i\) by \(1-U_i\) changes \(S_2\) to \(2-S_2\), so the density is symmetric around 1.

Three Uniform Variables: The Piecewise Quadratic Density

Now let \(S_3=U_1+U_2+U_3\). We can obtain its density by adding one more independent Uniform(0,1) variable to \(S_2\). The convolution integral is

where the second integral includes only the values for which \(s-u\in[0,2]\). The integration limits change at \(s=1\) and \(s=2\). Within the middle region, the integrand itself changes formula when \(s-u=1\), so that region must be split as well.

Here is the full calculation. For \(0<s<1\), the integration range is \(0\leq u\leq s\). Throughout this range \(0\leq s-u\leq1\), so \(f_{S_2}(s-u)=s-u\). Therefore,

For \(1<s<2\), the integration range is \(0\leq u\leq1\). When \(0\leq u\leq s-1\), \(s-u\geq1\), so the second piece of \(f_{S_2}\) applies. When \(s-1\leq u\leq1\), \(s-u\leq1\), so the first piece applies. Thus

Finally, if \(2<s<3\), the integration range is \(s-2\leq u\leq1\). Here \(1\leq s-u\leq2\), so the second piece of \(f_{S_2}\) applies throughout:

Outside \([0,3]\) there are no possible sums. The expressions agree at the breakpoints: both adjacent pieces equal \(1/2\) at \(s=1\), and both equal \(1/2\) at \(s=2\). This completes the derivation. \(\square\)

Checks and Worked Probabilities

A piecewise formula should be checked against the support and total probability, not just obtained algebraically. The three-variable density is nonnegative on each piece: on \(1\leq s\leq2\), its middle expression can be written as \( \frac34-(s-\frac32)^2 \), which ranges from \(1/2\) to \(3/4\). Its total integral is

The first and last pieces are mirror images around \(s=3/2\), as expected: replacing each \(U_i\) by \(1-U_i\) changes \(S_3\) to \(3-S_3\). These checks are useful for catching an incorrect sign or a missing piece.

Common Mistakes and Why the Shape Matters

These densities are useful both as concrete distributions and as models for understanding how adding independent variables changes a distribution. Even in this simple setting, the support and the density pieces must be tracked carefully.

• Assuming a sum of uniform variables is uniform. The density of \(S_2\) is proportional to the length of an overlap interval, not constant across its support. For example, \(f_{S_2}(0.2)=0.2\), while \(f_{S_2}(1)=1\).
• Using one formula across the whole support. The relevant overlap changes at 1 for \(S_2\), and at 1 and 2 for \(S_3\). The middle region for \(S_3\) also requires splitting its integral where \(s-u=1\).
• Confusing the density with a probability. A density value such as \(f_{S_3}(1.5)=0.75\) is not the probability that \(S_3=1.5\). For a continuous random variable, probabilities are integrals over intervals, and a single point has probability zero.
• Forgetting the support. The two-variable sum cannot be outside \([0,2]\), and the three-variable sum cannot be outside \([0,3]\). The density is zero elsewhere.

The piecewise forms arise because the set of inputs that can produce a given sum expands, then contracts, as the sum moves through its support. For two variables, the overlap length is linear. Adding a third variable integrates that linear density, producing quadratic pieces. The breakpoints mark where the geometry of the contributing region changes.